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An ordinary linear differential equation of first order has the form
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(1) |
where $y$ means the unknown function, $P$ and $Q$ are two known continuous functions.
For finding the solution of (1), we may seek a function $y$ which is product of two functions:
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(2) |
One of these two can be chosen freely; the other is determined according to (1).
We substitute (2) and the derivative $\frac{dy}{dx} = u\frac{dv}{dx}+v\frac{du}{dx}$ in (1), getting $u\frac{dv}{dx}+v\frac{du}{dx}+Puv = Q$ , or
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(3) |
If we chose the function $v$ such that $$\frac{dv}{dx}+Pv = 0,$$ this condition may be written $$\frac{dv}{v} = -P\,dx.$$ Integrating here both sides gives $\ln{v} = -\int P\,dx$ or $$v = e^{-\int Pdx},$$ where the exponent means an arbitrary antiderivative of $-P$ . Naturally, $v(x) \neq 0$ .
Considering the chosen property of $v$ in (3), this equation can be written $$v\frac{du}{dx} = Q,$$ i.e. $$\frac{du}{dx} = \frac{Q(x)}{v(x)},$$ whence $$u = \int\frac{Q(x)}{v(x)}\,dx+C = C+\!\int Qe^{\int Pdx}dx.$$
So we have obtained the solution
![$\displaystyle y = e^{-\int P(x)dx}\left[C+\!\int Q(x)e^{\int P(x)dx}dx\right]$](http://images.planetmath.org:8080/cache/objects/8717/js/img4.png "$\displaystyle y = e^{-\int P(x)dx}\left[C+\!\int Q(x)e^{\int P(x)dx}dx\right]$") |
(4) |
of the given differential equation (1).
The result (4) presents the general solution of (1), since the arbitrary constant $C$ may be always chosen so that any given initial condition $$y = y_0 \quad \mathrm{when}\quad x = x_0$$ is fulfilled.
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